### Discussion :: Probability - General Questions (Q.No.4)

Malasrvizhi said: (Oct 31, 2010) | |

How did get the sum element? |

Rama said: (Feb 14, 2011) | |

How to select n(e) = 4. |

Vian said: (Feb 17, 2011) | |

@rama. That is because the event of finding the sum 9 is 4. |

Azam said: (Apr 22, 2011) | |

In two throws of a dice, n(S) = (6 x 6) = 36. HOW IT DO PLEASE EXPLAIN |

Krish said: (May 16, 2011) | |

How it could be n(s)= (6 x 6) |

Srinivas said: (Jun 15, 2011) | |

n(s)=36 n(a)=[36][45][63][54] p(a)=4/36=1/9 |

Sreejith said: (Jun 17, 2011) | |

@Azam In both throughs, it can be any of 1,2,3...,6(6C1 ways of selection). So total number of outcomes = 6C1*6C1 = 36 More clearly, the elements of sample spaces are {(1,1),(1,2),(1,3),....,(2,1),(2,2),.....(6,1),(6,2),....(6,6)} Hope you got it. |

Harikannan said: (Jun 30, 2011) | |

How to find event that is (3, 4) (4, 5) (5, 4) (6, 3) please explain. |

Syam said: (Jan 5, 2012) | |

{ (3, 6) , (4, 5) , (5, 4) , (6, 3) } is a case only. We can get (6, 3) , (5, 4) , (3, 6) , (4, 5) also. So answer is 8/38 = 2/9. Am I correct ? |

Ankita said: (Feb 4, 2012) | |

How do we know that we have to pick up 6 * 6? |

Anurag said: (Mar 12, 2012) | |

Ankita:. Because 6c1 * 6c1 is 36. |

Kanchi said: (Sep 18, 2012) | |

s is a sample space i.e all possible outcomes if two coins are tossed then s={HH,TT,HT,TH} same in this qu. two dice -use then no of possible outcome s={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)}, {(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)} .......... {(6,1),(6,2),(6,3),(6,4),(6,5),(6,5)} n(s)=6+6+6+6+6+6=36 or 6*6=36 |

Frederick Mattson said: (Aug 6, 2013) | |

1,1 2,1 3,1 4,1 5,1 6,1 1,2 2,2 3,2 4,2 5,2 6,2 1,3 2,3 3,3 4,3 5,3 6,3 1,4 2,4 3,4 4,4 5,4 6,4 1,5 2,5 3,5 4,5 5,5 6,5 1,6 2,6 3,6 4,6 5,6 6,6 P(sum 9) = {(6,3),(5,4),(4,5),(3,6)}. = 4/36. = 1/9. |

Narottam said: (Oct 13, 2013) | |

Is there any short method to choose number of event ? |

Muhammad Rizwan said: (Oct 20, 2013) | |

I have a little bit confusion. Question is "What is the probability of getting a sum 9 from two throws of a dice?". Here we talk about two throws means that a dice is thrown 2 times. Further it means that a single dice is throw 2 times. So, we should take sample space of 6 because of a single dice. In case of 36 the question should be "What is the probability of getting a sum 9 when two dice are thrown?". Or perhaps I am in mistake in understanding the question. Please clear this. |

Dhaval said: (Nov 15, 2013) | |

If I three throw of dices than what is answer? |

Ranjith said: (Dec 21, 2013) | |

I understood of this way I don't know this is right or wrong. We want to getting some 36 = 36+9 = 45. 45+9 = 54. 54+9 = 63. |

Nitinpatel said: (Jan 20, 2014) | |

How do it? In two throws of a dice, n(S) = (6 x 6) = 36. |

Sus said: (Jul 30, 2014) | |

Is there any short method to choose number of event ? |

Purushotham said: (Aug 30, 2014) | |

Is there any short method to choose number of event ? |

Prajapati Divya H said: (Aug 31, 2014) | |

How do it? In throws of a dice, n(s) = (6*6) = 36. |

Sree Tulasi said: (Sep 23, 2014) | |

In that question they have asked for 2 dices, it means each dice is having 6 sides. So for 2 dices= 6*6=36. n(S)=36 Then we should find the possibility for sum of 9 so we can take (4,5),(3,6),(6,3),(5,4). These are 4 possibilities for sum of 9. n(E)=4. After that Probability of E = n(E)/n(S) = 4/36 = 1/9. |

Rama said: (Oct 5, 2014) | |

How we got n(e) i.e, 4. Please friends somebody help me to find out the answer. |

Lettisha L.S said: (Oct 19, 2014) | |

@Rama. So n(e)=4. It's because the possible combinations where you would get sum 9 when you throw the dice twice are 4. Like in a dice there are 1-6 no.s right? So what no.s do you put together to get sum 9? (3,6), (4,5), (6,3) ,(5,4) . But these combos aren't possible: (2,7), (1,8) because well they're not on a dice. Hope THAT cleared it up for you! ^_^. |

Nani said: (Feb 16, 2015) | |

How can we get outcomes as 36? Please explain. |

Ajax said: (Feb 19, 2015) | |

Short method of find sum of two dice. Sum - Time. 2 - 1. 3 - 2. 4 - 3. 5 - 4. 6 - 5. 7 - 6. 8 - 5. 9 - 4. 10 - 3. 11 - 2. 12 - 1. |

Reena said: (Apr 16, 2015) | |

Two of a dice n(S) = 6*6 because 1 dice = 1, 2, 3, 4, 5, 6. 1 dice = 1, 2, 3, 4, 5, 6. = 1(6)*1(6) = 36. Let E = Event of getting a sum = {(3, 6), (4, 5), (5, 4), (6, 3)}. = (3+6) = 9. = (4+5) = 9. = (5+4) = 9. = (6+3) = 9. As a question, P(E) = n(E)/n(S) = 4/36 = 1/9. |

John said: (Aug 13, 2015) | |

Do you need to do all thus working out? |

Laxmipriya said: (Sep 22, 2015) | |

Why to do multiplication like 6*6=36? Why we can't do addition? Like 6+6 = 12. I can't understand when to do addition and when to do multiplication. |

Amol Pawar said: (Oct 4, 2015) | |

Dice has 6 faces, i.e (1, 2, 3, 4, 5, 6) in this addition of 9 occurs when (5+4 or 4+5) and (6+3 or 3+6) that means 4 ways. In this way n(e) = 4 and n(s) = 6*6 =36. P (e) = n(e)/n(s) = 4/36 = 1/9. |

Deepthi said: (Mar 20, 2016) | |

By throwing 2 dies, to get a sum of 9 there are also possibilities of getting (1,8) (2,7) (3,6) (4,5) (5,4) (6,3) (7,2) (8,1) we have 8 possibilities please explain this. |

Naimulsamad said: (Jun 11, 2016) | |

Look in a dice there are number 1, 2, 3, 4, 5, 6. To form as a 9 we need 6 + 5 + 4 + 3 number. Such as 5 + 4 = 9, 6 + 3 = 9 but 1 + 8 or 2 + 7 is not possible because dice do not contain the number 7, 8. So sum up the number 6 +5 + 4 + 3 = 18 (one dice). 18 * 2 = 36 (two dice). |

Chethan said: (Jul 20, 2016) | |

Why they took n(E)/n(S)? In some other problem, they took n(S)/n(E). |

Umar said: (Oct 21, 2016) | |

9 + 9 = 18, => 2/18 = 1/9. |

B.Lokesh said: (Oct 31, 2016) | |

One dice means 6 faces. Two dice means 6 * 6 = 36 faces. Either, (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) (3,1)(3,2)(3,3)(3,4)(3,5)(3,6) (4,1)(4,2)(4,3)(4,4)(4,5)(4,6) (5,1)(5,2)(5,3)(5,4)(5,5)(5,6) (6,1)(6,2)(6,3)(6,4)(6,5)(6,6) Total = 36 outcomes. So question ask that, Sum =9. So, 3 + 6 = 9 (3,6) 6 + 3 = 9 (6,3) 4 + 5 = 9 (4,5) 5 + 4 = 9 (5,4) 4 out comes. p = num of outcomes/total outcomes. p = 4/36 = 1/9. Hope you understand this. |

Shankar said: (Nov 29, 2016) | |

Thanks for your solution @Sreejith. |

Harini said: (Dec 8, 2016) | |

How could n (S) be 36? |

Yallaling Dashavant said: (Jan 7, 2017) | |

Sum 9 means (1,8)(2,7)(3,6)(4,5)(5,4)(6 3)(7,2). Means 8/36 = 2/9. |

Sarita said: (Jan 27, 2017) | |

How can it be 6x6? Two throws of a dice mean 6+6. Your answer can only be till 12, not 36. |

Neetu Sanghvi said: (Feb 24, 2017) | |

I agree @Sree Tulasi. |

Prince said: (Jul 21, 2017) | |

Why we take these cases only? (3+6) = 9. (4+5) = 9. (5+4) = 9. (6+3) = 9. |

Pooja Gm said: (Apr 7, 2018) | |

Why we take this case only? (3+6) (4+5) (5+4) (6+3). This also possible; (7+2), (1+8), (2+7), (8+1), Please explain the reason. |

Swathi said: (May 11, 2018) | |

Why we should not use the permutations here?like sum 9 can appear in either (6, 3) or (3, 6) , (5, 4) or (4, 5). Why we only choose the one possibility? |

Joshua said: (Feb 7, 2019) | |

Because in a throw of two dices there is only 6x6, is the last the number won't Or should not exceed 6. |

Arif said: (Feb 15, 2019) | |

At first you draw sample space then you find pair which sum is 9, (1,1), (1,2),....... (5,4), (6,3),(3,6), (4,5). Total pair 36 and sum of 9 pair is 4 them p(sum 9)= 4 ÷ 36 = 1/9. |

Rrk said: (Feb 22, 2019) | |

Is throwing a dice two times and throwing two dice together, same? Please, anyone, clear me. |

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